Question

Prove that `(sqrt(3) + sqrt(2))^6 + (sqrt(3) - sqrt(2))^6` = 970

Answer 1

`(sqrt(3) + sqrt(2))^6 = ""^6"C"_0 (sqrt(3))^6 (sqrt(2))^0 + ""^6"C"_1 (sqrt(3))^5 (sqrt(2))^1 + ""^6"C"_2 (sqrt(3))^4 (sqrt(2))^2 + ""^6"C"_3 (sqrt(3))^3 (sqrt(2))^3 + ""^6"C"_4 (sqrt(3))^2 (sqrt(2))^4 + ""^6"C"_5 (sqrt(3))^1 (sqrt(2))^5 + ""^6"C"_6 (sqrt(3))^0 (sqrt(2))^6`

Since, ⁶C₀ = ⁶C₆ = 1, ⁶C₁ = ⁶C₅ = 6, ⁶C₂ = ⁶C₄ = `(6 xx 5)/(2 xx 1)` = 15, ⁶C₃ = `(6 xx 5 xx 4)/(3 xx 2 xx 1)` = 20

∴ `(sqrt(3) + sqrt(2))^6 = 1(27)(1) + 6(9sqrt(3))(sqrt(2)) + 15(9)(2) + 20(3sqrt(3))(2sqrt(2)) + 15(3)(4) + 6(sqrt(3))(4sqrt(2)) + 1(1)(8)`

∴ `(sqrt(3) + sqrt(2))^6 = 27 + 54 sqrt(6) + 270 + 120sqrt(6) + 180 + 24sqrt(6) + 8`    ...(i)

Also, `(sqrt(3) - sqrt(2))^6 = ""^6"C"_0 (sqrt(3))^6 (sqrt(2))^0 - ""^6"C"_1 (sqrt(3))^5 (sqrt(2))^1 + ""^6"C"_2 (sqrt(3))^4 (sqrt(2))^2 - ""^6"C"_3 (sqrt(3))^3 (sqrt(2))^3 + ""^6"C"_4 (sqrt(3))^2 (sqrt(2))^4 - ""^6"C"_5 (sqrt(3))^1 (sqrt(2))^5 + ""^6"C"_6 (sqrt(3))^0 (sqrt(2))^6`

= `1(27)(1) - 6(9sqrt(3)) (sqrt(2)) + 15(9)(2) - 20(3sqrt(3)) (2sqrt(2)) + 15(3)(4) - 6(sqrt(3)) (4sqrt(2)) + 1(1)(8)`

∴ `(sqrt(3) - sqrt(2))^6 = 27 - 54sqrt(6) + 270 - 120sqrt(6) + 180 - 24sqrt(6) + 8`   ...(ii)

Adding (i) and (ii), we get

`(sqrt(3) + sqrt(2))^6 + (sqrt(3) - sqrt(2))^6 = (27 + 54sqrt(6) + 270 + 120sqrt(6) + 180 + 24sqrt(6) + 8) + (27 - 54sqrt(6) + 270 - 120sqrt(6) + 180 - 24sqrt(6) + 8)`

= 54 + 540 + 360 + 16

= 970